Four-dimensional generalized difference matrix and some double sequence spaces

Abstract

In this study, I introduce some new double sequence spaces , , , and as the domain of four-dimensional generalized difference matrix in the spaces , , , and , respectively. I show that the double sequence spaces , and are the Banach spaces under some certain conditions. I give some inclusion relations with some topological properties. Moreover, I determine the α-dual of the spaces and , the -duals of the spaces , , , and , where , and the γ-dual of the spaces , and . Finally, I characterize the classes of four-dimensional matrix mappings defined on the spaces , , , and of double sequences.

On almost B-summable double sequence spaces

“…The four-dimensional generalized difference matrix and matrix domain of it on some double sequence spaces was recently defined and studied by Tuǧ and Başar [ 1 ], and Tuǧ [ 2 ]. The matrix was defined by for and for all .…”

“…In this paper, as natural continuation of [ 2 ] and [ 11 ], we introduce new almost null and almost convergent double sequence spaces and as the domain of four-dimensional generalized difference matrix in the spaces and , respectively. Throughout the paper, we suppose that the terms of the double sequence and are connected with equation ( 1.3 ) and the four-dimensional generalized difference matrix will be presented with .…”

“…Appl. 2017(1):149, 2017 ), we introduce new almost null and almost convergent double sequence spaces and as the four-dimensional generalized difference matrix domain in the spaces and , respectively. Firstly, we prove that the spaces and of double sequences are Banach spaces under some certain conditions.…”

On the domain of Riesz mean in the space Ls

“…In [14] Tuǧ and Başar have introduced some new double sequence spaces B(M u ), B(C ϑ ), and B(L s ) as the domain of four-dimensional generalized difference matrix B(r, s, t, u) in the spaces M u , C ϑ and L s , respectively.…”

“…x kl := s n (a m i n j kl ) , k = k i and l = l j , 0 , otherwise for all k, l ∈ N. Since s > 1, using the inequality (14) we see that…”